/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 A \(20.0 \mathrm{cm} \times 30.0... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 49

A \(20.0 \mathrm{cm} \times 30.0 \mathrm{cm}\) rectangular loop of wire carries \(1.0 \mathrm{A}\) of current clockwise around the loop. (a) Find the magnetic force on each side of the loop if the magnetic ficld is 2.5 T out of the page. (b) What is the net magnetic force on the loop?

Short Answer

Expert verified
Answer: The magnetic force on each side parallel to the magnetic field is 0 N, the magnetic force on each side perpendicular to the magnetic field is 0.75 N, and the net magnetic force on the loop is 1.5 N.

Step by step solution

01

Setting up the problem

Determine the lengths of the sides of the rectangular loop. Using the given dimensions, we know that the loop has two sides with lengths of 20.0 cm and two sides with lengths of 30.0 cm.
02

Calculate the magnetic force on the sides parallel to the magnetic field

Using the formula \(F = I\mathcal{L} \times B\), we calculate the magnetic force on the two sides with lengths of 20.0 cm: \(F_{1} = I\mathcal{L}_{1} \times B = (1.0\,\mathrm{A})(0.20\,\mathrm{m})(2.5\,\mathrm{T}) = 0.5\,\mathrm{N}\). For these sides, since the current is parallel to the magnetic field, there is no magnetic force acting on them, so \(F_{1} = 0\).
03

Calculate the magnetic force on the sides perpendicular to the magnetic field

Now, we calculate the magnetic force on the two sides with lengths of 30.0 cm: \(F_{2} = I\mathcal{L}_{2} \times B = (1.0\,\mathrm{A})(0.30\,\mathrm{m})(2.5\,\mathrm{T}) = 0.75\,\mathrm{N}\). For these sides, the current is perpendicular to the magnetic field, so the magnetic force is maximized.
04

Calculate the net magnetic force

We now have the magnetic forces for each of the sides: \(F_{1} = 0\) and \(F_{2} = 0.75\,\mathrm{N}\). Since the forces on the two sides parallel to the magnetic field are zero, the net magnetic force is the sum of the forces on the other two sides: \(F_{\mathrm{net}} = 2F_{2} = 2(0.75\,\mathrm{N}) = 1.5\,\mathrm{N}\). So, the magnetic force on each side parallel to the magnetic field is 0 N, the magnetic force on each side perpendicular to the magnetic field is 0.75 N, and the net magnetic force on the loop is 1.5 N.

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Most popular questions from this chapter

The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ A sample containing carbon (atomic mass 12 u), oxygen \((16 \mathrm{u}),\) and an unknown element is placed in a mass spectrometer. The ions all have the same charge and are accelerated through the same potential difference before entering the magnetic field. The carbon and oxygen lines are separated by \(2.250 \mathrm{cm}\) on the photographic plate, and the unknown element makes a line between them that is \(1.160 \mathrm{cm}\) from the carbon line. (a) What is the mass of the unknown clement? (b) Identify the element.
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